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Koenti007
New Member

Fastest way to join single property of object array into a string

Hi,

I have an object array eg:

[

{"title": "10","label": "Label10"},
{"title": "11","label": "Label11"}
]

 

What is the fastest way (less actions the better) to obtain a string variable composed of all Label properties eg 'Label10,Label11'?

Thx!

 

 

1 ACCEPTED SOLUTION

Accepted Solutions
Pstork1
Multi Super User
Multi Super User

The best I can come up with is 2 actions (assuming you've already done a Parse JSON)

1) A data Select statement to create an Array with only the Label properties in it

2) A Compose that uses three nested functions.  A Join to create a String from the Array, then a Replace to remove the beginning of the record and key, and finally a Replace to remove the trailing record indicator.

Replace(Replace(join(body('Select'),','),'{"label":',''),'}','')

image.png



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9 REPLIES 9
Paulie78
Super User
Super User

Try this:

Paulie78_0-1613431547887.png

 

Pstork1
Multi Super User
Multi Super User

The best I can come up with is 2 actions (assuming you've already done a Parse JSON)

1) A data Select statement to create an Array with only the Label properties in it

2) A Compose that uses three nested functions.  A Join to create a String from the Array, then a Replace to remove the beginning of the record and key, and finally a Replace to remove the trailing record indicator.

Replace(Replace(join(body('Select'),','),'{"label":',''),'}','')

image.png



-------------------------------------------------------------------------
If I have answered your question, please mark your post as Solved.
If you like my response, please give it a Thumbs Up.

Your answer could've been perfect if you click at this button (red circle). It will replace the interface from "key-value" pair to direct value instead. And you don't need the next Compose action.

I have just found out this.

 

2021-03-03_20-09-15.png

Paulie78
Super User
Super User

@radityoardi that’s what I showed in my answer. But perhaps I didn’t make it clear enough. 

That's the best and shortest solution.

Paulie78
Super User
Super User

Just for fun, as someone bumped the thread, here is an even shorter way, just one action:

faster.png

The expression in the compose is:

join(xpath(xml(json(concat('{ "root": {"Labels":', string(outputs('Array')), '}}'))), '//label/text()'), ',')
shahar
Frequent Visitor

the fastest way to create an array of strings is to use Select.

In the Select, you convert to text and use the expression: item().property.   Here is what it would look like

shahar_0-1677787605678.png

Here is what the data run looks like

shahar_1-1677787785951.png

 

 

 

IuriiPalchuk
Frequent Visitor

There is an easier way to achieve this using single Select action. 

 

Add new Select action, switch to advanced (Map) mode and insert Expression:

 

 

 

item()?['label']

 

 

 

IuriiPalchuk_0-1679400473693.png

This will result in a plain array of strings which you can easily join:

 

 

join(body('Select_Label'), ', ')

 

----------

Oops, the same idea appered twice in this thread even before me and I posted it one more time. 

This worked well for me, and saved me from an Apply to Each loop!

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